There are 10 Republicans, 8 Democrats and 2 Independent legislators eligible for committee membership.

How many 5-member committees exist which include at least two Republicans?

My Work :


$(\;All\; 5-member\; committees\;)\;-\;(\;committiees\; with \;no\; Republicans\;+\;committees\; with\; one \;Republican \;)$

Explanation: There are $C(20,5)$ 5-member committees which will be subtracted from sum of two numbers in order to achieve the answer; C(10,5) is number of 5-member committees which include no Republicans (I've only counted Democrats and Independents ($8+2$)then found number of 5-member committees ) and C(10,4) is number of 5-member committees with one Republican (Like before I've excluded Republicans but counted 4-member committees then added one Republican to each subset)

Book's Answer:

It's just said $C(10,2)C(18,3)$

  • 3
    $\begingroup$ The number of committiees with exactly one Republican is $10\cdot C(10,4)$ (there are $10$ ways to choose this Republican). This does not match book's answer, though. $\endgroup$ – ajotatxe Sep 15 '18 at 6:38
  • $\begingroup$ I'm not sure about that because , there was no mention of difference between each Republican $\endgroup$ – Abbas Sep 15 '18 at 6:44
  • 1
    $\begingroup$ The book answer is wrong, it's an overcount. Use the same logic for 2 R, 1 D, 2-member committee with at least 1 R. Answer is 3, but book would give 2-choose-1 times 2-choose-1, which is 4. $\endgroup$ – Gerry Myerson Sep 15 '18 at 6:50
  • $\begingroup$ What text is this? Please notify the publisher of the error. $\endgroup$ – N. F. Taussig Sep 15 '18 at 7:18
  • $\begingroup$ Link to the text ufile.io/mnowz page 15 or 16 $\endgroup$ – Abbas Sep 15 '18 at 7:25

I think that the answer is $$\binom{20}5-\binom{10}5-10\binom{10}4=13,152$$

and that book's answer is wrong. The idea behind book's answer seems to be the following:

First, we choose two Republicans. There are $C(10,2)$ ways to do it. Then we choose three any other members. There are $C(18,3)$ ways to do it. The problem is that then we are double (or even triple, etc) counting many possible choices. For an example, let's say that the Republicans are numbered from $1$ to $10$, and the other ones from $11$ to $20$. If we choose for first $1$ and $2$ and then $3$, $4$ and $12$ it is the same as choosing for first $3$ and $4$ and then $1$, $2$ and $12$.

Note (just in case you don't know): $\binom ab$ is just another way (and more usual, if you ask me) to write $C(a,b)$.

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  • $\begingroup$ Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least" $\endgroup$ – Abbas Sep 15 '18 at 7:05
  • $\begingroup$ @Abbas That would not explain the error. If there were exactly two Republicans, the count would be $\binom{10}{2}\binom{10}{3}$. $\endgroup$ – N. F. Taussig Sep 15 '18 at 7:08
  • $\begingroup$ @N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10? $\endgroup$ – Abbas Sep 15 '18 at 7:22
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    $\begingroup$ @Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $\binom{10}{2}\binom{10}{3}$. $\endgroup$ – N. F. Taussig Sep 15 '18 at 7:25
  • $\begingroup$ @N. F. Taussig Got it ! Thanks $\endgroup$ – Abbas Sep 15 '18 at 7:28

Republicans-$10$,Democrats-$8$,Independent legislator-$2$

There should be at least $2$ Republicans in the $5$ members committee.

So, there may be $2$ or $3$ or $4$ or $5$ republicans in the committee, as there are no restrictions on the numbers of the Democrats & Independent legislator in the committee-so we can treat these two groups as a single one of $8+2=10$ members.Call this new group the D-IL group.

The number of ways of forming such committee is

[($2$ Republicans & $3$ D-IL)or ($3$ Republicans & $2$ D-IL)or ($4$ Republicans & $1$ D-IL)or ($5$ Republicans & $0$ D-IL)]

=($2$ Republicans)$\cdot$ ($3$ D-IL)+($3$ Republicans)$\cdot$ ($2$ D-IL)+($4$ Republicans)$\cdot$ ($1$ D-IL)+($5$ Republicans)$\cdot$($0$ D-IL)

=${10 \choose 2}\cdot {10 \choose 3}+{10 \choose 3}\cdot {10 \choose 2}+{10 \choose 4}\cdot {10 \choose 1}+{10 \choose 5}\cdot {10 \choose 0}=13152.$

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  • $\begingroup$ Thank you for answering ! $\endgroup$ – Abbas Sep 15 '18 at 7:38

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